Question

Calculate the energy change by 23g of water vapor at converted to ice at $-38^{\circ}C$ .

145°O

Given:

Specific heat (ice) = 2.10 J/g* $^{\circ}C$

Specific heat (water) = 4.184 J/g* $^{\circ}C$

Specific heat (water vapor) = 1.996 J/g* $^{\circ}C$

        $\Delta H$ _fus = 333 J/g

       $\Delta H$ _vapor = 2260 J/g

Answer 1

Answer

Energy change = - 73.17kJ

Explanation

Heat released when temperature of water vapor decreased from 145°C to 100°C is q₁

q₁ = m × $\Delta$ T × C

= 23g ×(- 45°C) × 1.996J/g °C

= - 2.07kJ

Heat released when water vapor condense to liquid water is q₂

q₂ = - ( $\Delta$ H_vap × m)

= - (2260J/g × 23g)

= - 51.98kJ

Heat released when temperature of liquid water decreased from 100°C to 0°C is q₃

q₃ = m × $\Delta$ T × C

= 23g × - 100°C× 4.184J/g °C

= - 9.62kJ

Heat released when liquid water froze at 0°C is q₄

   q₄ = - ( $\Delta$ H_fus× m)

= - ( 333J/g × 23g)

= - 7.66kJ

Heat released when temperature of ice decreased from 0°C to - 38°C(q₅)

q₅ = m × $\Delta$ T × C

= 23g × - 38°C × 2.10J/g °C

= - 1.84kJ

Total heat change(q) = q₁ + q₂ + q₃  + q₄  + q₅

q = - 2.07kJ + (-51.98kJ) + ( - 9.62kJ) + (-7.66kJ) + (1.84kJ)

= - 73.17kJ